answer%20HW5 - HW 5 Solutions Problem 1 Taking the z- axis...

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Unformatted text preview: HW 5 Solutions Problem 1 Taking the z- axis to be the direction of the electric field and using z = r cos θ , the only non-zero transition matrix element is between the ground state and the 2 p state with m l = 0. Hence, H 210 , 100 ( t ) =- eE τ 2 + t 2 Z drψ * 210 ( r ) r cos θψ 100 ( r ) =- eE τ 2 + t 2 1 4 √ 2 πa 3 Z 2 π dφ Z π dθ cos 2 θ sin θ Z ∞ dr r 3 r a exp[- 3 r/ 2 a ] =- eE a τ 2 + t 2 2 7 √ 2 3 5 (1) The transition probability is thus P = e 2 E 2 a 2 ¯ h 2 2 15 3 10 Z ∞-∞ dt exp(- ωt ) τ 2 + t 2 2 = π 2 e 2 E 2 a 2 ¯ h 2 τ 2 2 15 3 10 exp(- 2 ωτ ) (2) where ω = 3 me 4 8(4 π ¯ h ) 2 (3) Problem 2 We denote the vacuum state by | i and excited states by | n i . These are the usual energy eigenkets of the 1D SHO. We note that h n | x 2 | i = h n ¯ h 2 mω a 2- + a + a- + a- a + + a 2 + i = ¯ h 2 mω δ n + √ 2 δ n 2 (4) where ω is the classical angular frequency. Using (1), and to first-order in time-dependent pertur- bation theory, c (1) n ( t ) =- i ¯ h Z t e inω t h n | H | i dt =- i ¯ h K Z t e inω t e- t /τ ¯ h 2 mω δ n + √ 2 δ n 2 dt (5) Thus, according to first-order theory, it can only transit to n = 2. c (1) 2 ( t ) =- i √ 2 K 2 mω Z t e 2 iω t- t /τ dt =- iK √ 2 2 mω " e (2 iω- 1 /τ ) t- 1 2 iω- 1 τ # (6) For t τ | c (1) 2 ( t ) | 2 ≈ K 2 2 m 2 ω 2 1 4 ω 2 + 1 /τ 2 (7) 1 Problem 3 The Hamiltonian reads: H = β [ S 1 x S 2 x + S 1 y S 2 y + S 1 z S 2 z ] . (8) In the basis set of | + + i , | - -i , | +-i , | - + i , the matrix representation of the Hamiltonian H ij = h i | H | j i is H = β ¯...
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This note was uploaded on 11/21/2010 for the course PHYSICS 137B taught by Professor Moore during the Spring '07 term at University of California, Berkeley.

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answer%20HW5 - HW 5 Solutions Problem 1 Taking the z- axis...

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